 # Evaluate the integral without using tables. \int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}} BenoguigoliB 2021-11-08 Answered
Evaluate the integral without using tables.
${\int }_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}}$
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Step 1
Given,
$I={\int }_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}}$
Step 2
Formula used:
$\int \frac{dx}{\sqrt{1-{x}^{2}}}={\mathrm{sin}}^{-1}\left(x\right)+C$
$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$
Step 3
Apply the above formula, we get
$I={\int }_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}}$
$={\left[{\mathrm{sin}}^{-1}\left(x\right)\right]}_{0}^{1}$
$={\mathrm{sin}}^{-1}\left(1\right)-{\mathrm{sin}}^{-1}\left(0\right)$
$={\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{\pi }{2}\right)\right)-{\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(0\right)\right)$
$=\frac{\pi }{2}-0$
$=\frac{\pi }{2}$